Apollonius <Pergaeus>; Lawson, John - The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

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94[17] and P will be greater than one to R and half AE, and of courſe, AQ
(a fourth proportional to S, R and P) greater than a third proportional
to P and half AE; in which caſe the rectangle AQ, P will be greater
than the ſquare on half AE, and ſo AD (a mean proportional between
AQ and P) greater than half AE; but when this happens, it is plain that
DH can neither cut nor touch the circle on AE, and therefore, the
problem becomes impoſſible.

PROBLEM III. (Fig. 4. and 5.)

In any indefinite ſtraight line let there be aſſigned the points A and E;
it is required to cut it in another point O, ſo that the ſquare on the ſegment
AO may be to the rectangle contained by the ſegment EO and a given line
P, in the ratio of two given ſtraight lines R and S.

Analysis. Suppoſe the thing done, and that O is the point ſought:
then will the ſquare on AO be to the rectangle EO, P as R to S. Make
AQ to P as R is to S; then will the ſquare on AO be to the rectangle EO,
P as AQ is to P; or (Eu. V. 15.) the ſquare on AO is to the rectangle EO,
P as the rectangle AQ, AO is to the rectangle P, AO; wherefore AO is to
EO as AQ to AO; conſequently by compoſition, or diviſion, AO is to AE
as AQ is to OQ, and ſo (Eu. VI. 16.) the rectangle AO, OQ is equal to
the rectangle AE, AQ; and hence, as the ſum or difference of AO and OQ
is alſo given, theſe lines themſelves are given by the 85th or 86th of the
Data.

Synthesis. Take AQ a fourth proportional to S, P and R, and
deſcribe thereon a circle; erect at A, the indefinite perpendicular AK, and
take therein AD, a mean proportional between AE and AQ; from D,
draw DH, parallel to AE, if O be required beyond E; but through F the
center of the circle on AQ, if it be ſought beyond A, or between A and
E, cutting the ſaid circle in H: Laſtly, from H draw HO perpendicular
to DH, which will cut the indefinite line in O, the point required.

For it is plain from the Conſtruction, that AD and HO are equal; and
(Eu. VI. 17) the rectangle AE, AQ is equal to the ſquare on AD, and
therefore equal to the ſquare on HO; but the ſquare on HO is equal to the
rectangle AO, OQ, (Eu. III. 35. 36) conſequently the rectangle AO,

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